The axle ratio is 4.10 with 15" rims..not worried about the tire height right now.


so with the exact same tire height, but a larger rim size of 1" greater, gives a loss of 1 tenth on the final drive ratio? Assuming all data is same, as about.


Example.

15" rim stock + 4.10 final drive ratio.
16" rim + 4.10 final drive ratio, is now what? 4.00????

Am I doing this right?

Thank you for the help members.

 

If you haven't changed the tire diameter, it doesn't matter what your rim size is. You are not changing your effective gearing (The relationship of mph to rpm's will be the same).

 

^+1

BTW, where did you get that conversion factor from?

 

Is the torque increased with a smaller rim size.

example:

Tire A is 225/50/15 on 15" rim
Tire B is 225/50/16 on 16" rim
Tire C is 225/50/14 on 14" rim...Is rim C applying the greatest amount of torque to the pavement?


Am I off base here?


I wasn't worried about speed or RPM at given setting. or the changes thereof

I was wondering, does a larger rim, with a tire that fits the larger rim, given all other factors are the same, create more momentum?.. Or does it create less momentum.

Assuming all weight for all examples are exactly the same...I know its probadly not possible, but its a idea I have, and I have noticed a big difference in the performance on a timed 1/4 track. Just seems to depend on the rim/tire size, considering all other aspects have not changed..


Am I crazy for asking this? I always thought the rear axle ratio was a direct factor in the amount of revolutions passed out to the rear wheels..thus increasing TQ or decreasing it..Is this wrong? if it is or isn't please explain fully, if possible... Thank you Club Members.


PS: This odd thought came from me and my crazy brain..

 

It's simple really, the wheels are the last gearing ratio. Tire size itself doesn't matter for now. It's the overall diameter that is important.

It is defined as rim diameter, 16" for example, plus the height of the tire. On a 225/50/16 size tire the height is defined as 50% of the width (225) in mm. So 225 x 0.5 = 112.5 mm / 25.4 = 4.43 inches. Total diameter is then 16 + 4.43 = 20.43 ".

An higher diameter will create an higher ratio, giving you more top speed but slower acceleration. On the opposite having a smaller overall diameter will help acceleration as it will create a lower ratio, essentially improving the torque sent to the ground.

Simply put a larger wheel/tire combination will have a bigger circonference, thus will cover more distance each turn of the wheel/tire. This result as more speed at any given rpm. But since the engine as to push an higher ration it loses power in the process.

Now let's get more complex. If you increase the diameter to much the engine will eventually lack enough power to push the car any faster because of wind resistance. This means that in practice you will not necessarily increase your top speed. On longer tracks however (No street racing/speeding please) this will allow you a higher speed in each gears which can be an advantage.

If you reduce your total wheel diameter it will provide a lower ratio to the engine and result in more acceleration but be aware this isn't as drastic changing the differential or transmission ratios. Downside is lower top speed in each gear and faster shifting.

Now let's get back to the tires themselves. Changing to a larger wheel with lower profile tires can have both avantages and disavantages. You get a stiffer sidewall which allows better traction in tight turns but you have to major downsides. First you will normally had weight to your wheels as they will be larger. I know race wheels are light but they aren't as light as smaller race wheels. Second you will move that weight toward the outside of the wheel. Both means that the wheels/tires gain more inertia and are harder to accelerate/stop.

Last thing, changing the diameter of your wheels will affect your speedo and it will need a recalibration.

In the end the tire and wheel combination you choose will make a significant difference in your car. I think I got pretty lost in the subject but I hope someone can find this useful lol.

 

Wow

 

wut?

 

Wow as in this entire post. Thats all I'll say on that matter, but you did make a mistake RB eater. You forgot that a wheel has TWO side walls withen its overall diameter, so add another 4.43 to your overall wheel diamter (225/50-16) for a total diamter of 24.86".

~Mike.............

 

Tire/wheel diameter vs. speedometer calculator: http://www.miata.net/garage/tirecalc.html

This goes hand in hand with what's already been pointed out about acceleration vs. top speed.

 

Tire manufacturers have already thought about this.

For a 15" wheel, 205 size tire, you'll most likely see 205/60/15
When you look for a 205 size tire in 16", you'll most likely see 205/55/16.

Tire manufacturers realize that as long as total circumference (and by extension, diameter) does not change, your speedo / Final drive ratio will be unchanged as well. Therefore, they work by downsizing the aspect ratio to compensate for the increased rim height.

As mentioned before:

the 15 inch wheel / tire combination total diameter is 15 inches + 2 X (205mm * 60% * mm to inch conversion factor). Why 2 times? The aspect ratio only accounts for the sidewall height, the distance from wheel circumference to tire circumference. But since drawing a line from one end of the tire/wheel combination to the other requires going into 2 sections of tire and one section of wheel, you have to take into account the sidewall height twice.

the mm to inch conversion factor is (1/25.4) or the reciprocal of 25.4

The total diameter for the 15 inch wheel/tire combination is 24.68 inches.

Calculating for a 16 inch wheel and a 205/55/16 tire, you arrive at a diameter of 24.88 inches. While not exactly the same, the size is virtually unchanged. If you're really nitpicky, there's a difference of about 0.8% when changing between the sizes, which you're probably not going to notice.

Similarly, for a 225 size tire, you'll most likely see tires in the following sizes: 225/50/16, 225/45/17 and 225/40/18. 25" is a very common diameter.

Personally, on an 5.12 R&P NA track prepped car, I'd run a 15" x 8" wheel with a 225/45/15 tire. The total diameter is closer to 23" now, which has the effect of using a higher final drive ratio. My effective wheel torque advantage as compared to a 25" wheel/tire car with a standard 4.1 final drive and the same amount of power is now: (25/23)*(5.12/4.1) = 36% improvement in acceleration and a 36% penalty in top speed, but since FC's are geared to over 200 mph in 5th gear at 8000 rpm, the top speed penalty isn't really that worrisome. If I build my engine to withstand 10000 rpm's, I'll gain back some of that top speed, reducing my penalty to 29% as compared to a standard 8000 rpm engine car, with a 25" diameter and a 4.1 R&P gear.

 

So a smaller rim/tire setup accels faster, is less weight, but gives less traction, during starts.. It also has a lower top speed.

Larger rim/tire setup is slower to accelerate from stop, weights more, and give higher top speed. Provides more traction from stop.

Is there any way to work out how much weight per say each rim/tire combo applys the greatest amount of traction with, and when its not enough weight without changing tire brands or things of that nature.?

 

I would think traction is dependent on tire width, where acceleration and top speed is dependent on total tire/wheel diameter.

For example, there are 275 mm width tires on a 15 inch rim (275/35/15) which has a total diameter of 22.58". It'll accelerate a lot faster than a normal 25" diameter tire. It'll also provide loads of traction based on it's width. Sure it'll have more rolling resistance than a tire that's narrower, but for the purposes of gearing changes, we can ignore that fact since that applies anytime you move to a wider tire.

You may want to rephrase that question, it's a little confusing to understand.

 

Let me try to explain.

The n/a 13b has little TQ to use. Smaller rim/tire combo seems to pull harder from a start, spins more, even spins in 3rd gear.

Larger rim/tire combo spins less, seems slower from start, gets a faster recorded 1/4 track time.

I understand its because less loss of traction because the applied TQ to the ground is less..right? thus makes quicker accel, over a greater distance. vs a smaller rim/tire setup.

But the question is 14" rim combo seems to apply gobs of TQ to the track. is that just from smaller rim/tire size, or weight of it?

what would be faster on recorded track time slicks of 14" rims or 16" rims with slicks?

sorry, I am not sure, if I have asked this right yet...damnit..

 

Don't think of torque as usable power. Only horsepower is that and we have the same amount no matter the gear ratio or wheel diameter. Think of torque at the ground as leverage. That's really what it is. On trains they call this tractive effort which is really nothing more than torque at the rails. A higher tractive effort is more leverage with which to use your power. Obviously more low end tractive effort brings a tradeoff of less on the top end but if it's only acceleration to a certain speed that you're looking for, this isn't an issue.

 

^
your still here!
I bet you dont remember me. But you came by SGP that one time when I had my N/a motor apart.